Keean Schupke k.schupke at imperial.ac.uk
Thu Aug 18 09:14:22 EDT 2005

```Christian Maeder wrote:

>Keean Schupke wrote:
>
>
>>implementation of unify? For example can the algorithm be simplified
>>from my nieve attempt? Most importantly is it correct?
>>
>>
>
>It will not be correct without occurs check. You may also get different
>terms for the same variable in your substitution list.
>
>
Prolog does not use an occurs check... and this is embedding prolog.
However I accept this point, thats why there was the comment about
no occurs check in the code. I actually want to cope with recursive
definitions as Prolog does, so the solution to:

X = f(X)

should be:

f(f(f(f(...))))

which is an infinite recursion.

>The simplest form of unification does not take a substitution as input
>and uses functions to compose two substitutions and to apply a
>substitution to a term.
>
>
>
>>   unify :: Subst -> (Term,Term) -> [Subst]
>>
>>
This signature came from the paper... The input subst is an accumulator
and it would normally be Id when calling - so there is effectively no input
substitution.

>
>Do you ever get real lists? The result type "Maybe Subst" is more
>appropriate.
>
>
No, I dont think the algorithm gives real lists, Maybe would be better,
although I
think I will get it working before playing with changing the rest of the
code.
Is it possible to ever have more than one meaningful answer from
unification?

>
>
>>   unify' s [] [] = [s]
>>   unify' s (t0:ts) (u0:us) = case unify s (t0,u0) of
>>           s@(_:_) -> unify' (concat s) ts us
>>           _ -> []
>>
>>
>
>input lists of different lengths should not cause a runtime error but
>only a unification failure (indicated by "Nothing" or "[]" in your case.)
>
>
>
Aha, a genuine bug... thanks!

>HTH Christian
>
>Here's a part of my version:
>
>unify' (t0:ts) (u0:us) = do
>   s1 <- unify (t0,u0)
>   s2 <- unify' (map (applySubst s1) ts)
>                (map (applySubst s1) us)
>   return (composeSubst s1 s2)
>
>
I am now using:

unify' :: Subst -> [Term] -> [Term] -> [Subst]
unify' s (t0:ts) (u0:us) = case unify s (t0,u0) of
s@(_:_) -> unify' (concat s) ts us
_ -> []
unify' s [] [] = [s]
unify' _ _ _ = []

Regards,
Keean.
```